A smoker carries two match boxes -- one in his right pocket, and one in his left

Question

A smoker carries two match boxes -- one in his right pocket, and one in his left pocket. Initially, there are 10 matches in each of the match boxes. Each time he wants to light a cigarette, with probability .6 he will take a match from his right pocket, and with probability .4 he will take a match from his left pocket. At some point he reaches for his right pocket and finds out that the matchbox is empty. What is the probability that at that moment, there are exactly 4 matches in the left pocket?


Hint: Identify the relevant random variable.

Explanation / Answer

In order for this to happen, he must have previously drawn 16 matches and taken 10 from the right and 6 from the left. This is a binomial random variable X with n = 16, p = 0.6, q = 0.4, where X represents the number of times he chose the right pocket, so: P(X=10) = (16C10) (0.6)^10 (0.4)^4 = 0.198

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